Question: $\overline{AB}$ = $3\sqrt{10}$ $\overline{BC} = {?}$ $A$ $C$ $B$ $3\sqrt{10}$ $?$ $ \sin( \angle BAC ) = \frac{3\sqrt{10} }{10}, \cos( \angle BAC ) = \frac{ \sqrt{10}}{10}, \tan( \angle BAC ) = 3$
Solution: $\overline{AB}$ is the hypotenuse $\overline{BC}$ is opposite to $\angle BAC$ SOH CAH TOA We know the hypotenuse and need to solve for the opposite side so we can use the sine function (SOH) $ \sin( \angle BAC ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\overline{BC}}{\overline{AB}}= \frac{\overline{BC}}{3\sqrt{10}} $ $ \overline{BC}=3\sqrt{10} \cdot \sin( \angle BAC ) = 3\sqrt{10} \cdot \frac{3\sqrt{10} }{10} = 9$